[Guest]

In the park one day, I saw a great assortment of children and dogs. Counting heads, I saw 22. Counting legs, I saw 68.

How many children and dogs were in the park?

[Guest]

  Reveal hidden contents

12 dogs 10 people

WAY TOO EASY

[unreality]

  guest_7314 said:

In the park one day, I saw a great assortment of children and dogs. Counting heads, I saw 22. Counting legs, I saw 68.

How many children and dogs were in the park?

  Reveal hidden contents

c = children, d = dogs

c+d = 22

2c+4d = 68

subtract the first from the second:

(2c+4d)-(c+d) = 68-22

c+3d = 46

subtract the first from this

(c+3d)-(c+d) = 46-22

2d = 24

d = 12

which means c has to be 22-12 = 10

10 children, 12 dogs

checking: 20 legs + 48 legs = 68 legs

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alternatively, you could do it this way:

c+d = 22

2c+4d = 68

divide the second equation by 2

c+d = 22

c+2d = 34

subtract the first from the second, and you get d = 12, and it follows that c = 10

[Guest]

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12 dogs= 48 heads

+10 people=20 heads

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22 heads =68 heads

[Guest]

  Reveal hidden contents

Correct; there were 10 children and 12 dogs.